Theorem sbel2x 2299

Description: Elimination of double substitution. (Contributed by NM, 5-Aug-1993.) (Proof shortened by Wolf Lammen, 29-Sep-2018.)

Assertion

Ref	Expression
sbel2x	|- ( ph <-> E. x E. y ( ( x = z /\ y = w ) /\ [ y / w ] [ x / z ] ph ) )

Distinct variable group:   x, y, ph

Allowed substitution hints:   ph( z, w)

Proof of Theorem sbel2x

Step	Hyp	Ref	Expression
1		nfv 1772	. . 3 |- F/ y ph
2		nfv 1772	. . 3 |- F/ x ph
3	1, 2	2sb5rf 2291	. 2 |- ( ph <-> E. y E. x ( ( y = w /\ x = z ) /\ [ y / w ] [ x / z ] ph ) )
4		ancom 456	. . . 4 |- ( ( y = w /\ x = z ) <-> ( x = z /\ y = w ) )
5	4	anbi1i 706	. . 3 |- ( ( ( y = w /\ x = z ) /\ [ y / w ] [ x / z ] ph ) <-> ( ( x = z /\ y = w ) /\ [ y / w ] [ x / z ] ph ) )
6	5	2exbii 1730	. 2 |- ( E. y E. x ( ( y = w /\ x = z ) /\ [ y / w ] [ x / z ] ph ) <-> E. y E. x ( ( x = z /\ y = w ) /\ [ y / w ] [ x / z ] ph ) )
7		excom 1938	. 2 |- ( E. y E. x ( ( x = z /\ y = w ) /\ [ y / w ] [ x / z ] ph ) <-> E. x E. y ( ( x = z /\ y = w ) /\ [ y / w ] [ x / z ] ph ) )
8	3, 6, 7	3bitri 279	1 |- ( ph <-> E. x E. y ( ( x = z /\ y = w ) /\ [ y / w ] [ x / z ] ph ) )

Syntax hints:   <-> wb 189   /\ wa 375   E.wex 1674   [wsb 1808

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1680  ax-4 1693  ax-5 1769  ax-6 1816  ax-7 1862  ax-10 1926  ax-11 1931  ax-12 1944  ax-13 2102

This theorem depends on definitions:  df-bi 190  df-an 377  df-ex 1675  df-nf 1679  df-sb 1809

This theorem is referenced by: (None)